You have 6 columns, not 5. It would be clearer to say what A is and what b is separately.
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I find it hard to understand what you're trying to do. Are you trying to solve the linear system Ax=b (where I assume b is the rightmost column)?
If so, then you will, of course, have a lot more than just one solution, since the rank of your matrix is 2, meaning you have the freedom to choose values for 3 out of 5 coordinates of the x vector.
In any case, I would find the general solution for the homogeneous system Ax=0 (a linear combination of a 3 vector base {v1,v2,v3} of solutions to the system, since your matrix has rank 2), as well as a private solution to the system Ax=b (say Xp), and conclude that the solution set for your equation is {Xp}+Sp{v1,v2,v3}.
In your particular case, if you choose your base to be v1=(2,1,0,0,0), v2=(3,0,2,-1,0) and v3=(1,0,4,0,-1), and take for example Xp=(0,1,5,0,0), then you conclude your solution set is {(0,1,5,0,0)}+Span{(2,1,0,0,0),(3,0,2,-1,0),(1,0,4,0,-1)} (you may confirm it for yourself).
Forgive my poor english.
Hi, I had a couple of problems with some matrices and I'd appreciate any help if possible:
Ax=b in row echelon form, need to find all solns:
1 -2 0 3 1 -2
0 0 1 2 4 5
0 0 0 0 0 0
0 0 0 0 0 0
the 5th column (-2,5,0,0) is the solution set. Forgive my lack of decent syntax. So I was thinking maybe I could start with x5=a, x4=b, x3=5-a-b, or what? I'm suspicious that that I might get too many variables, although I think there are infinitely many solutions.
Also if I have a 5 by 3 matrix such that b=a1+a2=a2+a3 i am supposed to see what this tells me about the number of solutions to this Ax=b. I was hoping to write out something like b=x1(a11+a21+...+a51)+x2(a12+...+a52)+x3(a13+a23+...+a53) to try and get an expression with a11+a12+a13+a14+a15=a1, etc. to work with but I really don't know what I'm supposed to be doing.